what is the derivative of ln (5/n) could u show me step by step?

Answer 1

y=ln(5/n)=ln(5) -ln(n) (according to logarithm rules)
dy/dn = -1/n

Answer 2

The derivative of ln(z) is 1/z*dz. Use the substitution rule and let z = 5/n. Then dz = -5/n^2. Substitute back in and the derivative of ln(5/n) is 1/(5/n)*-5/(n^2) = -1/n.

Answer 3

Remember that the derivative of ln(x) = 1/x.
However this question has 5/n instead of x so you need to use the chain rule.

the derivative of ln(5/n) = 1/(5/n) * the deriviative of 5/n
the derivative of ln(5/n) = 1/(5/n) * -5/n^2
the derivative of ln(5/n) = n/5 * -5/n^2
the derivative of ln(5/n) = -1/n

Alternatively you could firstly use your log laws and rewrite the question.

ln(5/n) = ln(5) - ln(n)
And then take the derivative. Note ln(5) is a constant so its derivative is zero.
So it is just the deriviative of -ln(n) which is -1/n.

Answer 4

let f(n)= ln (5/n)

differentiating,

f'(n)= d/dn(ln (5/n) )
= 1/(5/n) * d/dn( 5/n)
= n/5 * 5 * d/dn(n^-1) [ d/dx (x^a) = ax^(a-1) ]
= n * (-1) n^(-1-1)
= -n* 1/n^2
= -1/n (ans)

Answer 5

sure!

let y = ln(5/u)

you need to use the chain rule:
d/dx f(g(x)) = f'(g(x)) * g'(x)

let u = 5/u
then d/du ln(u) = 1/u

5/n = 5n^-1

d/dx (u)^n = n * u^(1-n)

d/dn (5n^-1) = -5n^-2 = -5/n^2

y' = 1/(5/n) * -5/n^2

y' = n/5 * -5/n^2

y' = -1/n <== answer

Answer 6

y = ln ( 5/n )
Let u = 5 / n = 5 n^ ( -1 )
du/dn = ( - 5 ) n^( -2 ) = ( - 5 ) / n ²
dy/du = 1 / u
dy/dn = (dy/du) (du/dn)
dy/dn = ( 1 / u ) ( - 5 ) / n ²
dy/dn = (n / 5) ( - 5 ) / n ²
dy/dn = (- 1 / n )